Question 613983
{{{x + sqrt(x) + 5 = 7}}}
The easy way, IMHO, to solve this is to recognize that since the exponent of x, 1, is twice the exponent of {{{sqrt(x)}}}, 1/2, this equation is in what is called "quadratic form". Quadratic form equations can be solved like quadratic equations.<br>
If you are new to equations of quadratic form equations, then it can be very helpful to use a temporary variable:
Let q = {{{sqrt(x)}}}
Then {{{q^2 = (sqrt(x))^2 = x}}}
Substituting these into our equation we get:
{{{q^2 + q + 5 = 7}}}
This is obviously a quadratic equation. To solve, we first make one side equal to zero. Subtracting 7 from each side we get:
{{{q^2 + q - 2 = 0}}}
Then we factor:
(q + 2)(q - 1) = 0
One of these factors must be zero:
q + 2 = 0 or q - 1 = 0
Solving these we get:
q = -2 or q = 1<br>
But we are not interested in what values "q" might be. We want to know values for "x". So we substitute back in for the q's:
{{{sqrt(x) = -2}}} or {{{sqrt(x) = 1}}}
Since square roots are never negative there is no solution to the first equation. But we will get solution from the second equation. Squaring both sides we get:
x = 1<br>
Since we squared both sides, we <i>must</i> check our solution. Use the original equation to check:
{{{x + sqrt(x) + 5 = 7}}}
Checking x = 1:
{{{(1) + sqrt((1)) + 5 = 7}}}
{{{1 + 1 + 5 = 7}}}
7 = 7 Check!<br>
After a few of these quadratic form equations, you will no longer need a temporary variable. You will see that
{{{x + sqrt(x) - 2 = 0}}}
will factor into:
{{{sqrt(x) + 2)(sqrt(x)-1) = 0}}}
etc.<br>
An alternate way to solve this is as a square root equation:
{{{x + sqrt(x) + 5 = 7}}}
1. Isolate a the square root
Subtracting x and 5 from each side:
{{{sqrt(x) = 2-x }}}
2. Square both sides:
{{{(sqrt(x))^2 = (2-x)^2 }}}
{{{x = 4 - 4x + x^2}}}
3. Solve the resulting equation.
This is quadratic so we want one side to be zero, Subtracting x from each side:
{{{0 = 4 - 5x + x^2}}}
4. Factor (or use the Quadratic Formula):
(x - 4)(x - 1) = 0
5. One factor must be zero:
x - 4 = 0 or x - 1 = 0
6. Solve
x = 4 or x = 1
7. Check. (Again, we squared both sides earlier so the check is not optional!)
{{{x + sqrt(x) + 5 = 7}}}
Checking x = 4:
{{{(4) + sqrt((4)) + 5 = 7}}}
4 + 2 + 5 = 7
11 = 7 Check failed! Reject x = 4
Checking x = 1:
{{{(1) + sqrt((1)) + 5 = 7}}}
{{{1 + 1 + 5 = 7}}}
7 = 7 Check!